(Long answer, but I had no choice...)
Once again, before doing anything else, you must determine what the concentration of the dissolved carbon dioxide will be
CO_(2(g)) rightleftharpoons CO_(2(aq))
This is done by using Henry's law, which implies that
P_("gas") = k_H * C_("gas"), where
P_("gas") - the partial pressure of the gas;
k_H - Henry's constant - it has a specific value for every gas and it's temperature-dependent;
C_("gas") - the concentration of the dissolved gas.
Since you're at "25"^@"C", the value of Henry's constant is listed at "29.41 L atm/mol", which means that (again, convert mbar to atm)
C_(CO_2) = P_(CO_2)/k_H = "0.00691 atm"/"29.41 L atm/mol" = "0.000235 mol/L"
You can now use this value to determine the given reaction's equilibrium constant
CO_(2(aq)) + H_2O_((l)) rightleftharpoons H_((aq))^(+) + HCO_(3(aq))^(-)
K_c = ([H^(+)] * [HCO_3^(-)])/([CO_2]) = (0.0000398 * 0.0117)/(0.000235) = 0.00198
(I've converted all the given concentrations into "mol/L").
Now you add the extra H^(+) to the equilibrium. The new concentration of H^(+) will be
H_("new")^(+) = H^(+) + H_("added")^(+) = 39.8mumol + 79.6 mumol = 119.4mumol
Since the concentration of one of the species involved in the equilibrium changes, you can calculate the reaction quotient for that change
Q_c = ([H_("new")^(+)] * [HCO_3^(-)])/([CO_2]) = (0.000119 * 0.0117)/(0.000235) = 0.00592
The first observation you can make is that Q_c is biger than K_c, which is in accordance with the fact that we now have more products than reactants; as a result the equilibrium will shift to the left and favor the formation of more reactants.
The Gibbs free energy change for a system that has a reaction quotient Q_c is
DeltaG_("rxn") = DeltaG^@ + RT*ln(Q_c)
Since at equilibrium DeltaG^@ = -RT * ln(K_c), the above equation becomes
DeltaG_("rxn") = RT * ln(Q_c/K_c)
Therefore,
DeltaG_("rxn") = "8.314 J/mol K" * "298.15 K" * ln(0.00592/0.00198)
DeltaG_("rxn") = "2715 J/mol" = "+2.72 kJ/mol"
A positive value for DeltaG_("rxn") means that the forward reaction will not be spontaneous; however, the reverse reaction will be spontaneous.
